Question

87.
In the figure, AP : PC = 3 : 4, BM : MP = 3 : 2 and BQ = 12cm find
4×AQ/7​

Answer 1

We have ratios of sides, So The Best and Easy way is to use properties of lines.

Construct : Extend P to meet AQ in Z such that,

PZ || CQ.

So, AQ = AZ + QZ

Now, In △PZB,

We have, PZ || QM,

By Basic proportionality theorem,

$\boxed{ \frac{BM}{MP} = \frac{BQ}{QZ}}$

From the question, BM : MP = 3 : 2

BQ = 12 cm.

$\boxed{ \frac{BM}{MP} = \frac{BQ}{QZ}} \\ \\ \frac{3}{2} = \frac{12}{QZ } \\ \\ QZ = \frac{12 \times 2}{ 3} \\ \\ QZ = 8cm$

Now, In △AQC,

We have PZ || QC,

By Basic proportionality theorem,

$\boxed{ \frac{AZ}{QZ}= \frac{AP}{PC}}$

From the question,

AP : PC = 3 : 4

We already found QZ = 8cm.

$\boxed{ \frac{AZ}{QZ}= \frac{AP}{PC}} \\ \\ \frac{AZ}{8} = \frac{3}{4} \\ \\ AZ = \frac{3}{4} \times 8 \\ \\ AZ = 6 \: cm$

Now, AQ = AZ + QZ = 6 + 8 = 14 cm

Required Answer

4/7 × AQ = 4/7 × 14 = 8